1) The document discusses the Cauchy Integral Theorem and Formula. It states that if a function f(z) is analytic inside and on a closed curve C, then the integral of f(z) around C is equal to 0. 2) It provides examples of evaluating integrals using the Cauchy Integral Theorem when the singularities lie outside the closed curve C. 3) The Cauchy Integral Formula is introduced, which expresses the value of an analytic function F(a) inside C as a contour integral around C. Examples are worked out applying this formula to find the value and derivatives of functions at points inside C.

1. Subject :- Complex Variable & Numerical Method
(2141905)
Topic :- Cauchy Integral Theorem & Formula
Branch :- Mechanical
Semester :- 4th
•Made By :-
-Divyangsinh Raj (150990119004)
-Naved Fruitwala (150990119006)
-Utkarsh Gandhi (150990119007)
Guided By :-
Dr. Purvi Naik

2. Cauchy Integral Theorem
• If f(z) is analytic function and f '(z) is
continuous at each point inside and on
closed curve c then,
 
c
0dz)z(f

3. Examples
1. Evaluate where c is circle |z| = 2.
f (z) is not analytic at z = - 4. Here c is |z| = 2
which is a circle with center (0,0) and
radios 2.
 c
4z
dz
4z
1
)z(f



4. Here z = - 4 lies outside the curve c. Therefore
f'(z) is analytic inside and on c and also f '(z) is
continuous in z plane.
Therefore by Cauchy integral theorem,
0
4z
dz
c



5. 2. Evaluate ; |c| : |z-1|=1
c is circle with center (1,0) and radios 1.
f(z) is not analytic at z=-2i
For |z-1|= |-2i-1 |
 c
z
i2z
e
15
14



6. z = - 2i lies outside to curve c. So f(z) is
analytic on and inside c. f '(z) is continuous on
and inside c. Therefore by Cauchy integral
theorem,
 
c
z
0
i2z
e

7. 3. Evaluate ; |c| : |z|=1
f (z) is not analytic at cos z = 0.
i.e.
Now c is circle with center (0,0) and radios 1.
These points lie outside c. So the f(z) is analytic and
f(z) is continuous. So by Cauchy integral theorem,
c
zdzsec
zcos
1
zsec)z(f 
,....
2
5
,
2
3
,
2
z






0zdzsec
c


8. Cauchy Integral Formula
• If a function F(a) is analytic inside a closed
curve C and if a is any point inside C then,
• 𝐹 𝑎 =
1
2𝜋𝑖
𝐹(𝑥)
𝑧−𝑎
. 𝑑𝑧

9. Cauchy Integral Formula For
Derivative
• If a function F(x) is analytic in region R then its
derivative at any point z=a is also analytic in R
and it is given by
• 𝐹′ 𝑎 =
1
2𝜋𝑖
𝐹(𝑥)
(𝑧−𝑎)2 . 𝑑𝑧
• In general,
• 𝐹 𝑛
𝑎 =
𝑛!
2𝜋𝑖
𝐹(𝑥)
(𝑧−𝑎) 𝑛+1 . 𝑑𝑧

10. Ex – 1 :-
𝒛
𝒛−𝟐
. 𝒅𝒛 , where C is 𝒛 − 𝟐 =
𝟑
𝟐
• Solution :-
• Here I =
𝑧
𝑧−2
. 𝑑𝑧
• And C is a circle with center (2,0) and radius
3
2
• Here z=2 lies inside C, so let F(z) = z
• F(z) is analytic everywhere inside and on C,
• Now, by Cauchy Integral Formula,
• 𝐹 𝑎 =
1
2𝜋𝑖
𝐹(𝑍)
(𝑍−𝑎)
. 𝑑𝑧

11. Cont...
•
𝐹(𝑧)
(𝑧−𝑎)
. 𝑑𝑧 = 2𝜋𝑖 𝐹(𝑎)
•
𝑧
𝑧−2
. 𝑑𝑧 = 2𝜋𝑖 𝐹(2)
•
𝑧
𝑧−2
. 𝑑𝑧 = 2𝜋𝑖 2
•
𝑧
𝑧−2
. 𝑑𝑧 = 4𝜋𝑖

12. Ex – 2 :-
𝒅𝒛
(𝒛 𝟐−𝟕𝒛+𝟏𝟐)
, C is |Z| = 3.5
• Let 𝐼 =
𝑑𝑧
(𝑧2−7𝑧+12)
• 𝐼 =
𝑑𝑧
(𝑧−4)(𝑧−3)
• 𝐼 =
1 (𝑧−4)
(𝑧−3)
. 𝑑𝑧
• Here, C is a circle with center (0,0) and radius 3.5
• Z=3 lies inside C, so let F(z) = 1 (𝑧 − 4)
• F(z) is analytic everywhere inside and on C,
• Now, by Cauchy Integral Formula,

13. Cont...
•
𝐹(𝑧)
(𝑧−𝑎)
. 𝑑𝑧 = 2𝜋𝑖 𝐹(𝑎)
•
𝑑𝑧
(𝑧2−7𝑧+12)
=
2𝜋𝑖 𝐹(3)
•
𝑑𝑧
(𝑧2−7𝑧+12)
=
2𝜋𝑖
1
3−4
•
𝑑𝑧
(𝑧2−7𝑧+12)
= −2𝜋𝑖

14. Ex – 3 :-
𝒔𝒊𝒏𝝅𝒛 𝟐+𝒄𝒐𝒔𝝅𝒛 𝟐
𝒛−𝟏 (𝒛−𝟐)
. 𝒅𝒛 where C is |z|
= 3
• Here, C is a circle with (0,0) and radius 3
• Now,
1
𝑧−1 (𝑧−2)
=
𝐴
𝑧−1
+
𝐵
(𝑧−2)
• By partial fraction,
• 1 = 𝐴 𝑧 − 2 + 𝐵 𝑧 − 1
• For 𝑧 = 2, 𝐵 = 1
• For 𝑧 = 1, 𝐴 = −1
•
𝑠𝑖𝑛𝜋𝑧2+𝑐𝑜𝑠𝜋𝑧2
𝑧−1 (𝑧−2)
. 𝑑𝑧 = −
𝑠𝑖𝑛𝜋𝑧2+𝑐𝑜𝑠𝜋𝑧2
𝑧−1
. 𝑑𝑧 +
𝑠𝑖𝑛𝜋𝑧2+𝑐𝑜𝑠𝜋𝑧2

15. Cont...
• Here, C is a circle with center (0,0) and radius
3
• Z=1 and 2 lies inside C, so let F(z) = 𝑠𝑖𝑛𝜋𝑧2
+
𝑐𝑜𝑠𝜋𝑧2
• F(z) is analytic everywhere inside and on C,
• Now, by Cauchy Integral Formula,
• 𝐹 𝑎 =
1
2𝜋𝑖
𝐹(𝑧)
(𝑧−𝑎)
. 𝑑𝑧

16. Cont...
• 𝐼 = −2𝜋𝑖 𝐹 1 +
2𝜋𝑖 𝐹(2)
• 𝐼 = −2𝜋𝑖 0 − 1 +
2𝜋𝑖 0 + 1
• 𝐼 = 2𝜋𝑖 + 2𝜋𝑖
• 𝐼 = 4𝜋𝑖

17. Ex – 4 :-
𝐳
(𝐳−𝟏) 𝟑 . 𝐝𝐳, C is |z|=2
• Let 𝐼 =
z
(z−1)3 . dz
• 𝐼 =
z
(z−1)2+1 . dz
• Here F(z)=z and a=1 which is on C
• Here, C is a circle with center (0,0) and radius
2
• F(z) is analytic everywhere inside and on C,
• Now by Cauchy Integral Formula For
Derivative,

18. Cont...
•
z
(z−1)3 . dz = 𝐹′′ 1 ×
2𝜋𝑖
2!
• 𝐹 𝑧 = 𝑧
• 𝐹′ 𝑧 = 1
• 𝐹′′ 𝑧 = 0
•
z
(z−1)3 . dz = 0 × 𝜋𝑖
•
z
(z−1)3 . dz = 0